0
$\begingroup$

Here's the formula...

I'm having some trouble using the time dilation formula. Say an astronaut leaves Earth for 10 years, at 0.85c. How much time has passed according to an observer on Earth?

I tried using the following formula:

$$t = \frac{1}{\sqrt{1-(v^2/c^2)}}$$

but couldn't seem to get an answer that made sense.

Any help would be much appreciated! This concept is super confusing to me.

$\endgroup$
9
  • $\begingroup$ I think you might be missing a term in your formula. $\endgroup$
    – Kyle Kanos
    Commented Sep 10, 2014 at 18:40
  • 2
    $\begingroup$ Leaves for 10 years in what frame? $\endgroup$
    – BMS
    Commented Sep 10, 2014 at 18:41
  • $\begingroup$ Okay. I wrote the formula exactly how it appeared on my formula sheet, but it's possible I'm missing something. Any idea what term? $\endgroup$
    – McB
    Commented Sep 10, 2014 at 18:42
  • $\begingroup$ Leaves for 10 years relative to the astronaut $\endgroup$
    – McB
    Commented Sep 10, 2014 at 18:44
  • $\begingroup$ Your image equation is a little different from the formula you typed, do you see what is missing? $\endgroup$
    – Kyle Kanos
    Commented Sep 10, 2014 at 19:33

1 Answer 1

5
$\begingroup$

Your written text says "t = 1/sqrt[1-(v^2/c^2)]". If you used that equation, it's no wonder you got a nonsensical value. In the equation in the scanned image, that's a letter $t$ in the numerator, not the digit $1$.

Also, although it works for this problem, the scanned image should really say something like

$$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}\ .$$

As it's written, it looks like it's expressing a coordinate transformation between the two frames instead of just expressing the time dilation factor, and interpreted as a coordinate transformation the equation would in general be wrong. The general coordinate transformation for the standard configuration is

$$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \left( t - \frac{vx}{c^2} \right)\ .$$

That equation reduces to the scanned equation if you're only dealing with the world line $x=0$. $x=0$ in this problem expresses that the astronaut is standing still in his coordinate system.

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.